Check my logic: Does a generalized dihedral extension necessarily contain the extension element? (I say yes.)

Question

I want to make sure I am correct about something I read in Moore and Pollatsek's "Difference Sets":

Suppose we have an abelian group $H$. If $\exists g\notin H$ with $g^2=1$ and $ghg^{-1}=h^{-1}\;\forall h\in H$ such that a group $G$ can be written as $H+gH$ in the integral group ring $\mathbb{Z}G$, then $G$ is called a generalized dihedral extension of $H$.

The question arises, since they don't make it explicit in the definition, whether $g\in G$, necessarily. I think this is true.

It certainly can be, for they give the clear example in the text where $H=\langle a,b|a^6=b^2=1, ab=ba\rangle$ and $G=\langle a,b,c| a^6=b^2=c^2=1, ab=ba, ac=ca^{-1}, bc=cb\rangle$ is a dihedral extension of $H$ where $c$ is the extending element.

I think it is necessary that $g\in G$ because if $G=H+gh\in\mathbb{Z}G$, then the elements of $G$ are precisely those of $H$ and those of the form $gh$ with $h\in H$. Since $H$ must have an identity, there is an $h\in H$ such that $gh=g$ and so $g\in G$. I don't know why I am doubting myself, but could someone confirm? I'll sleep better.

Perhaps there is an easier way to see that $g\in G$? Or if I'm wrong, an easy counterexample?

Answer 1

This is a terribly presented definition. A much more reasonable definition would be as follows:

Let $H$ be an abelian group. A group $G$ is the GDE of $H$ if it is the external semidirect product of $H$ with $C_2$, where conjugation by the non-identity element acts as the inverse map on $H$. That is, it is the group formed by $G=H\rtimes_\phi C_2$ where $\phi_c(h)=chc^{-1}=h^{-1}$ with $c$ being the non-identity element of $C_2$.

You can see this definition being used here and here. With this rephrasing, the book's definition's mention of $\mathbb{Z}G$ doesn't make a whole lot of sense (or maybe it does, it's been a while since I've done this) but the rest of what is said does. Why they chose to write it like that escapes me.

So yes, $g\in G$. In fact, $g\in G$ is crucial to the construction of $G$ in the first place! It's the image of the non-identity element of $C_2$ when mapped into $G$ as a subgroup in this construction.

Note that the motivation for this is that if you let $H=C_n$ then you get the GDE of $H$ is $D_{2n}$, so this is an extension of a group to look like a dihedral group.